Multi part upload

[Sushil Sureka <su...@gmail.com>]

Before I ask my question, I just wanted to let you know that I have
been through FAQs and other places but I can't seem to get it right. I
have this piece of code to upload the file

           File file = new File("c:\\temp.txt");
            FilePart filePart = new FilePart("theFile", file);
            try {
              ByteArrayOutputStream os = new ByteArrayOutputStream();
              filePart.send(os);
              InputStream i = convert(os);              
              request.setUserData(i);
              //request.setContentType("multipart/form-data");
              i.close();
            } catch (IOException e) {
              System.out.println("Error Building Multipart");
            }

     public static InputStream convert( ByteArrayOutputStream os )
throws IOException
     {    
         return new ByteArrayInputStream(os.toByteArray()); 
     }   

But when the app tried to parse the data on server side, it gets this exception
Failed to parse multipart request
org.apache.commons.fileupload.FileUploadException: the request was rejected beca
use no multipart boundary was found

I am using http client 2.0.2

Any help will be appreciated.
thanks
Sushil

Re: Multi part upload

[Kazuhito SUGURI <su...@lab.ntt.co.jp>]

Hi Sushil,

In article <c8...@mail.gmail.com>,
Thu, 31 Mar 2005 14:46:29 -0600,
Sushil Sureka <su...@gmail.com> wrote: 
sushil> Before I ask my question, I just wanted to let you know that I have
sushil> been through FAQs and other places but I can't seem to get it right. I
sushil> have this piece of code to upload the file
sushil> 
sushil>            File file = new File("c:\\temp.txt");
sushil>             FilePart filePart = new FilePart("theFile", file);
sushil>             try {
sushil>               ByteArrayOutputStream os = new ByteArrayOutputStream();
sushil>               filePart.send(os);
sushil>               InputStream i = convert(os);              
sushil>               request.setUserData(i);
sushil>               //request.setContentType("multipart/form-data");
sushil>               i.close();
sushil>             } catch (IOException e) {
sushil>               System.out.println("Error Building Multipart");
sushil>             }
sushil> 
sushil>      public static InputStream convert( ByteArrayOutputStream os )
sushil> throws IOException
sushil>      {    
sushil>          return new ByteArrayInputStream(os.toByteArray()); 
sushil>      }   
sushil> 
sushil> But when the app tried to parse the data on server side, it gets this exception
sushil> Failed to parse multipart request
sushil> org.apache.commons.fileupload.FileUploadException: the request was rejected beca
sushil> use no multipart boundary was found

You might refer RFC1867 'Form-based File Upload in HTML' to know
how file-upload works.
	http://www.ietf.org/rfc/rfc1867.txt


The RFC shows an example of data sent back from client as follows:
        Content-type: multipart/form-data, boundary=AaB03x
 
        --AaB03x
        content-disposition: form-data; name="field1"
 
        Joe Blow
        --AaB03x
        content-disposition: form-data; name="pics"; filename="file1.txt"
        Content-Type: text/plain
                                                                                
         ... contents of file1.txt ...
        --AaB03x--

As this example shows, you need to specify a boundary string
in the Content-Type field for the HTTP request,
'Content-type: multipart/form-data, boundary=YOUR-OWN-BOUNDARY'
for example. Then, use the boundary string to separate and terminate
each part of the multi-part content.

Hope this helps,
----
Kazuhito SUGURI
mailto:suguri.kazuhito@lab.ntt.co.jp